In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0), and C(c,0) is there. If the angle from a point A is 90o, what is the value of c?
A. 16/5
B. 12/5
C. 9/5
D. 7/5
E. 3/5
*An answer will be posted in 2 days.
In the x-y plane, a triangle ABC connected by 3 points A(0,4
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- Max@Math Revolution
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The line segment connecting point B with point A has a slope of 4/5: (y2 - y1) / (x2 - x1) = (4 - 0) / (0 - (-5)) = 4/5.Max@Math Revolution wrote:In the x-y plane, a triangle ABC connected by 3 points A(0,4), B(-5,0), and C(c,0) is there. If the angle from a point A is 90o, what is the value of c?
A. 16/5
B. 12/5
C. 9/5
D. 7/5
E. 3/5
*An answer will be posted in 2 days.
Since we are told that Angle A is a right angle, this means that the line segment connecting points B and C must be perpendicular to the line segment connecting points A and B. To be perpendicular, the slope of this second line must be the negative reciprocal of line segment AB. Thus, line segment BC has slope of -5/4.
Armed with this knowledge, and the fact that the y-coordinate of point C is 0, we can simply set up a ratio and solve for x.
We know that from point A to point C we go down four units, thus -5/4 = -4/x.
Solving for x: x = 16/5. Answer choice A.
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The slope of the line that passes through the points A and B is =(4-0)/(0-(-5))=4/5.
The slope of the line that passes through the points A and C is (4-0)/(0-c)=-4/c.
As they meet perpendicular, the product of the slope is -1. So, (4/5)(-4/c)=-1. So, we get 16/5c=1, c=16/5. Hence, the correct answer is A.
The slope of the line that passes through the points A and C is (4-0)/(0-c)=-4/c.
As they meet perpendicular, the product of the slope is -1. So, (4/5)(-4/c)=-1. So, we get 16/5c=1, c=16/5. Hence, the correct answer is A.
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